1. Generating Functions I Great Theoretical Ideas In Computer Science V. Adamchik D. Sleator CS 15-251 Spring 2010 Lecture 6 Jan. 28, 2010 Carnegie Mellon University + + ( ) + ( ) = ? X1 X1 X2 X2 + + X3 X3

2. Could Your iPod Be Holding the Greatest Mystery in Modern Science? Warm-Up Question a

3. “ Computing (The Algorithm) will be the most disruptive scientific paradigm since quantum mechanics." B.Chazelle http://www.cs.princeton.edu/~chazelle/pubs/algorithm.html

4. It would have to be The Algorithm. Jan. 12, 2006, The Economist If Google is a religion, what is its God?

5. Plan Counting with generating functions Solving the Diophantine equations Proving combinatorial identities Applied Combinatorics, by Alan Tucker Generatingfunctionology, by Herbert Wilf

6. Counting with Generating Functions We start with George Polya’s approach (1887-1985), Hungarian mathematician.

7. Problem Ula is allowed to choose two items from a tray containing an apple, an orange, a pear, a banana, and a plum. In how many ways can she choose? 5 2

8. There is a correspondence between paths in a choice tree and the cross terms of the product of polynomials!

9. Counting with Generating Functions (0 apple + 1 apple)(0 orange + 1 orange) (0 pear + 1 pear)(0 banana + 1 banana) (0 plum + 1 plum) In this notation, apple 2 stand for choosing 2 apples., and + stands for an exclusive or. Take the coefficient of x 2.

10. Problem Ula is allowed to choose two items from a tray containing TWO apples, an orange, a pear, a banana, and a plum. In how many ways can she choose? The two apples are identical.

11. Counting with Generating Functions 11 (0 apple + 1 apple + + 2 apple) (0 orange + 1 orange)(0 pear + 1 pear) (0 banana + 1 banana)(0 plum + 1 plum) Take the coefficient of x 2.

12. The function f(x) that has a polynomial expansion f(x)=a 0 + a 1 x + …+ a n x n is the generating function for the sequence a 0, a 1, …, a n

13. If the polynomial (1+x+x 2 ) 4 is the generating polynomial for a k, what is the combinatorial meaning of a k ? The number of ways to select k object from 4 types with at most 2 of each type.

14. The 251-staff Danish party The 251-staff wants to order a Danish pastry from La Prima. Unfortunately, the shop only has 2 apple, 3 cheese, and 4 raspberry pastries left. What the number of possible orders?

15. Counting with Generating Functions (1+x+x 2 ) (1+x+x 2 +x 3 ) (1+x+x 2 +x 3 +x 4 ) apple cheese raspberry = 1+3x+6x 2 +9x 3 +11x 4 +11x 5 +9x 6 +6x 7 +3x 8 +x 9 The coefficient by x 8 shows that there are only 3 orders for 8 pastries. Note, we solve the whole parameterized family of problems!!!

16. The 251-staff wants to order a Danish pastry from La Prima. The shop only has 2 apple, 3 cheese and 4 raspberry pastries left. What the number of possible orders? Raspberry pastries come in multiples of two.

17. Counting with Generating Functions (1+x+x 2 ) (1+x+x 2 +x 3 ) (1+x 2 +x 4 ) apple cheese raspberry = 1+2x+4x 2 +5x 3 +6x 4 +6x 5 +5x 6 +4x 7 +2x 8 +x 9 The coefficient by x 8 shows that there are only 2 possible orders.

18. Problem Find the number of ways to select R balls from a pile of 2 red, 2 green and 2 blue balls Coefficient by x R in (1+x+x 2 ) 3

19. (1+x+x 2 ) 3 = (x 0 +x 1 +x 2 )(x 0 +x 1 +x 2 )(x 0 +x 1 +x 2 )

20. Find the number of ways to select R balls from a pile of 2 red, 2 green and 2 blue balls e 1 + e 2 + e 3 = R 0 b e k b 2

21. Exercise Find the number of integer solutions to x 1 +x 2 +x 3 +x 4 =21 0 b x k b 7 Take the coefficient by x 21 in (1+x+x 2 +x 3 +x 4 +x 5 +x 6 +x 7 ) 4

22. Exercise Find the number of integer solutions to x 1 +x 2 +x 3 +x 4 = 21 0 < x k < 7

23. x 1 +x 2 +x 3 +x 4 = 21 0 < x k < 7 Observe 0 < 0 < x k < 7 -1 < -1 < x k -1 < 6 0 b x k -1 b 5 Than (x 1 -1)+(x 2 -1)+(x 3 -1)+(x 4 -1) = 21-4

24. x 1 +x 2 +x 3 +x 4 =21 0<x k <7 The problem is reduced to solving y 1 + y 2 + y 3 + y 4 = 17, 0 b y k b 5 Solution: take the coefficient by x 17 in (1+x+x 2 +x 3 +x 4 +x 5 ) 4 which is 20

25. Problem x 1 +x 2 +x 3 =9 1 b x 1 b 2 2 b x 2 b 4 1 b x 3 b 4

26. Solution x 1 +x 2 +x 3 =9 1 b x 1 b 2 (x+x 2 ) 2 b x 2 b 4 (x 2 +x 3 +x 4 ) 1 b x 3 b 4 (x+x 2 +x 3 +x 4 ) Take the coefficient by x 9 in the product of generating functions.

27. Two observations 1. A generating function approach is designed to model a selection of all possible numbers of objects. 2. It can be used not only for counting but for solving the linear Diophantine equations

28. The 251-staff Danish party Adam pulls a few strings… and a large apple Danish factory is built next to the CMU.

29. The 251-staff Danish party. Unfortunately, still only 3 cheese and 4 raspberry pastries are available. Raspberry pastries come in multiples of two. What the number of possible orders?

30. Counting with Generating Functions (1+x+x 2 +x 3 ) (1+x 2 +x 4 ) (1+x+x 2 +x 3 + …) cheese raspberry apple There is a problem (…) with the above expression.

31. Counting with Generating Functions (1+x+x 2 +x 3 ) (1+x 2 +x 4 ) (1+x+x 2 +x 3 + …)= cheese raspberry apple (1+x+2x 2 +2x 3 +2x 4 +2x 5 +x 6 +x 7 ) (1+x+x 2 +x 3 + …) =1+2x+4x 2 +6x 3 +8x 4 +…

32. The power series a 0 + a 1 x + a 2 x 2 +… is the generating function for the infinite sequence a 0, a 1, a 2, …,

33. A famous generating function from calculus :

34. We can use the generating function technique to solve almost all the mathematical problems you have met in your life so far.

35. Exercise Count the number of N-letter combinations of MATH in which M and A appear with repetitions and T and H only once. Coefficient by x N in (1+x+x 2 +…) 2 (1+x) 2

36. What is the coefficient of X k in the expansion of: ( 1 + X + X 2 + X 3 + X 4 +.... ) n ? A solution to: e 1 + e 2 +... + e n = k e k ≥ 0

37. What is the coefficient of X k in the expansion of: ( 1 + X + X 2 + X 3 + X 4 +.... ) n

38. GF for the pirates and gold problem But what is the LHS?

40. GF for the pirates and gold problem

41. (when x ≠ 1) (when |x| < 1)

42. We will be dealing with formal power series (aka generating functions)

43. The coefficients a k can be any complex numbers, though we will be mainly working over the integers.

44. Power series have no analytic interpretation.

45. FORMAL MANIPULATION: We deal with power series as we deal we numbers!

46. FORMAL MANIPULATION: 1/P(X) is defined to be the polynomial Q(X) such that Q(X) P(X) = 1. Th: P(X) will have a reciprocal if and only if a 0  0. In this case, the reciprocal is unique.

47. Power series make a commutative ring!

48. FORMAL MANIPULATION: The derivative of is defined by:

49. Power Series = Generating Function Given a sequence of integers a 0, a 1, …, a n We will associate with it a function The On-Line Encyclopedia of Integer Sequences

50. Generating Functions Sequence: 1, 1, 1, 1, … Generating function:

51. Find a generating function f(x) 1, 2, 3, 4, 5, …

52. Generating Functions F 0 =0, F 1 =1, F n =F n-1 +F n-2 for n ≥ 2

53. = x (x+x 2 +2 x 3 +3 x 4 +5 x 5 +8 x 6 +13 x 7 +…)(1-x-x 2 ) = x + x 2 + 2 x 3 + 3 x 4 + 5 x 5 + 8 x 6 + 13 x 7 +… - x 2 - x 3 - 2 x 4 - 3 x 5 - 5 x 6 - 8 x 7 -… - x 3 - x 4 - 2 x 5 - 3 x 6 - 5 x 7 -…

54. If a sequence satisfies a linear recurrence relation, we can always find its generating function.

55. Use generating functions to show that Proving Identities with GF

56. Prove by induction Prove algebraically Prove combinatorially Prove by Manhattan walk (see previous lecture) 15-355 Computer Algebra

57. Consider the right hand side: Coefficient by x n

58. Compute a coefficient of x k Use the binomial theorem to obtain

59. Find a closed form Computing Combinatorial Sums

60. Don't try to evaluate the sum that you're looking at. Instead, find the generating function for it, then read off the coefficients. Computing Combinatorial Sums

61. Multiply it by x n and sum it over n Interchange the sums

62. Take r = n - k as the new dummy variable of inner summation We recognize the inner sum as x k (1 + x) k

63. This is a geometric series The RHS is our old friend …

64. What did we find?

66. Generating Functions It’s one of the most important mathematical ideas of all time!

67. Study Bee Counting with generating functions Solving the Diophantine equations Proving combinatorial identities